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Question
A string, fixed at both ends, vibrates in a resonant mode with a separation of 2⋅0 cm between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to 1⋅6 cm. Find the length of the string.
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Solution
Given:
Separation between two consecutive nodes when the string vibrates in resonant mode = 2.0 cm
Let there be 'n' loops and
\[\lambda\] be the wavelength.
∴
\[\lambda\] = \[2 \times Separation between the consecutive nodes\]
\[\lambda_1 = 2 \times 2 = 4 \text{ cm }\]
\[\lambda_2 = 2 \times 1 . 6 = 3 . 2 cm\]
Length of the wire is L.
In the first case:
\[L = \left( \frac{n \lambda_1}{2} \right)\]
In the second case:
\[L = \left( n + 1 \right)\frac{\lambda_2}{2}\]
\[ \Rightarrow \frac{n \lambda_1}{2} = \left( n + 1 \right) \frac{\lambda_2}{2}\]
\[ \Rightarrow n \times 4 = \left( n + 1 \right)\left( 3 . 2 \right)\]
\[ \Rightarrow 4n - 3 . 2n = 3 . 2\]
\[ \Rightarrow 0 . 8 n = 3 . 2\]
\[ \Rightarrow n = 4
\text{ ∴ length of the string,}\]
\[L = \frac{\left( n \lambda_1 \right)}{2} = \frac{\left( 4 \times 4 \right)}{2} = 8 \text{ cm }\]
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